Showing the polarization of (complex) quadratic form is sesquilinear.

Question

Let $q: V\times V \to \mathbb{C}$ on a complex vector space $V$ be a quadratic form. Define $\tilde q$ by the polarization identity:

$$ \begin{equation} \tilde q(\phi,\psi) = \frac{1}{4} [q(\phi + \psi) -q(\phi - \psi) + iq(\phi + i\psi) - iq(\phi - i\psi)] \end{equation} $$

My question is how can I show that $\tilde q$ is sesquilinear?

My plan is to show that (1) $\tilde q$ is skew-symmetric, and (2) linear in the second argument. However, to show (2), I'm not exactly sure how to do so given that $q$ is defined in complex space. I found a hint that tells me to show that

a. $\tilde q(\phi, 2\psi) = 2\tilde q(\phi, \psi)$
b. $\tilde q(\phi, \psi + \psi') = \tilde q(\phi, \psi) + \tilde q(\phi, \psi')$
c. $\tilde q(\phi, \pm i\psi ) = \pm i\tilde q(\phi, \psi)$
d. $\tilde q(\phi, \alpha\psi ) = \alpha\tilde q(\phi, \psi )$ for all dyadic rationals in $\mathbb C$.

It makes sense to me that we need to take the complex part into account for $c$ and $d$. However, I have no idea what $d$ means. Also for $a$, is the number 2 just some random real integer?

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PS: Here's the definition I want to use for (complex) quadratic form:

1. $q(\lambda x) = |\lambda|^2q(x)\quad \forall \lambda\in\mathbb C, x\in V$
2. $q(\phi+\psi) + q(\phi-\psi) = 2q(\psi)+2q(\phi)$

From 2, we could rewrite the polarization identity as

$$ \tilde q(\phi,\psi) = \frac{1}{2}[q(\phi+i\psi)-q(\phi)-q(\psi)] - \frac{i}{2}[q(\phi+i\psi)-q(\phi)-q(i\psi)] $$

But I'm unsure if that's helpful for showing $\tilde q$ is sesquilinear.

Thanks for the help!

Answer 1

I think the given hints are quite helpful. Nevertheless a complete proof is somewhat lengthy. Here we show part (a).

We consider $\tilde q: V\times V\to \mathbb{C}$ which fulfills the polarisation identity: \begin{align*} \tilde q(\phi,\psi) = \frac{1}{4} [q(\phi + \psi) -q(\phi - \psi) + iq(\phi + i\psi) - iq(\phi - i\psi)] \tag{1} \end{align*}

We obtain for $\phi,\psi,\xi\in V$: \begin{align*} \color{blue}{\tilde{q}}&\color{blue}{(\phi,\psi)+\tilde{q}(\phi,\xi)}\tag{2.1}\\ &=\frac{1}{4}\left(q(\phi+\psi)-q(\phi-\psi)+iq(\phi+i\psi)-iq(\phi-\psi)\right.\\ &\quad+\left.q(\phi+\xi)-q(\phi-\xi)+iq(\phi+i\xi)-iq(\phi-\xi)\right)\tag{2.2}\\ &=\frac{1}{4}\left(q\left(\left(\phi+\frac{\psi+\xi}{2}\right)+\frac{\psi-\xi}{2}\right) +q\left(\left(\phi+\frac{\psi+\xi}{2}\right)-\frac{\psi-\xi}{2}\right)\right.\\ &\quad-q\left(\left(\phi-\frac{\psi+\xi}{2}\right)-\frac{\psi-\xi}{2}\right) -q\left(\left(\phi-\frac{\psi+\xi}{2}\right)+\frac{\psi-\xi}{2}\right)\\ &\quad+iq\left(\left(\phi+i\frac{\psi+\xi}{2}\right)+i\frac{\psi-\xi}{2}\right) +iq\left(\left(\phi+i\frac{\psi+\xi}{2}\right)-i\frac{\psi-\xi}{2}\right)\\ &\quad\left.-iq\left(\left(\phi-i\frac{\psi+\xi}{2}\right)-i\frac{\psi-\xi}{2}\right) -iq\left(\left(\phi-i\frac{\psi+\xi}{2}\right)+i\frac{\psi-\xi}{2}\right)\right)\tag{2.3}\\ &=\frac{1}{2}\left(q\left(\phi+\frac{\psi+\xi}{2}\right)+q\left(\frac{\psi-\xi}{2}\right) -q\left(\phi-\frac{\psi+\xi}{2}\right)-q\left(\frac{\psi-\xi}{2}\right) \right.\\ &\quad\left.+iq\left(\phi-i\frac{\psi+\xi}{2}\right)+iq\left(\frac{\psi-\xi}{2}\right) -iq\left(\phi+i\frac{\psi+\xi}{2}\right)-iq\left(\frac{\psi-\xi}{2}\right)\right)\tag{2.4}\\ &\,\,\color{blue}{=2\tilde{q}\left(\phi,\frac{\psi+\xi}{2}\right)}\tag{2.5} \end{align*}

Putting $\xi=0$ in (1) we obtain \begin{align*} \color{blue}{\tilde{q}(\phi,0)}=\frac{1}{4} [q(\phi) -q(\phi) + iq(\phi) - iq(\phi)] \color{blue}{=0}\tag{2.6} \end{align*} Combining (2.1), (2.5) and (3) we finally obtain \begin{align*} \color{blue}{\tilde{q}(\phi,\psi)=2\tilde{q}\left(\phi,\frac{\psi}{2}\right)}\tag{2.7} \end{align*} and the hint (a) follows.

Comment:

• In (2.2) we use the polarisation identity (1) for both terms $\tilde{q}(\phi,\psi)$ and $\tilde{q}(\phi,\xi)$.

• In (2.3) we split the terms conveniently as preparation for the next step.

• In (2.4) we apply the parallelogram identity \begin{align*} q\left(\phi+\psi\right)+q\left(\phi-\psi\right)=2\left(q\left(\phi\right)+q\left(\psi\right)\right) \end{align*}

• In (2.5) we cancel terms and apply (1) again.

Note: This derivation follows closely section 1.2 of Linear Operator in Hilbert Spaces by Joachim Weidmann. The other parts can also be derived from this section.

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Add-on [2022-04-17]:

With respect to comments to this post here is some additional information. In order to show sesquilinearity of $\tilde{q}:V\times V\to \mathbb{C}$ we have to show for all $\phi,\psi,\xi\in V$ and for all $a,b\in\mathbb{C}$: \begin{align*} \color{blue}{\tilde{q}\left(a\phi+b\psi\xi\right)}&\color{blue}{=a\tilde{q}\left(\phi,\xi\right)+b\tilde{q}\left(\psi,\xi\right)}\tag{*}\\ \color{blue}{\tilde{q}\left(\phi,a\psi+b\xi\right)}&\color{blue}{=a^{*}\tilde{q}\left(\phi,\psi\right)+b^{*}\tilde{q}\left(\psi,\xi\right)}\tag{**}\\ \end{align*} which is linearity in the first component and antilinearity in the second component. We now list the steps which can be used to show (*) and (**) together with some details.

additivity second component

In order to show antilinearity in the second component we show at first additivity \begin{align*} \color{blue}{\tilde{q}\left(\phi,\psi+\xi\right)=\tilde{q}\left(\phi,\psi\right)+\tilde{q}\left(\phi,\xi\right)}\tag{3} \end{align*}

We obtain from (2.1) and (2.5) \begin{align*} \color{blue}{\tilde{q}(\phi,\psi)+\tilde{q}(\phi,\xi)}&=2\tilde{q}\left(\phi,\frac{\psi+\xi}{2}\right)\tag{$\to 2.5$}\\ &\,\,\color{blue}{=\tilde{q}(\phi,\psi+\xi)}\tag{$\to 2.7$} \end{align*} and additivity (3) follows.

additivity first component

We want to show \begin{align*} \color{blue}{\tilde{q}\left(\phi+\psi,\xi\right)=\tilde{q}\left(\psi,\xi\right)+\tilde{q}\left(\psi,\xi\right)}\tag{4.1} \end{align*} and do so by establishing at first \begin{align*} \color{blue}{\tilde{q}\left(\phi,\psi\right)^{*}=\tilde{q}\left(\psi,\phi\right)}\tag{4.2} \end{align*}

We obtain \begin{align*} \color{blue}{\tilde{q}(\phi,\psi)^{*}} &=\frac{1}{4}\left(q\left(\phi+\psi\right)-q\left(\phi-\psi\right) +i q\left(\phi+i\psi\right)-i q\left(\phi-i \psi\right)\right)^{*}\tag{$\to (1)$}\\ &=\frac{1}{4}\left(q\left(\phi+\psi\right)-q\left(\phi-\psi\right)\right)\\ &\qquad+\frac{1}{4}\left(-i q\left((i)\left(\psi-i\phi\right)\right)+i q\left((-i)\left(\psi+i \phi\right)\right)\right)\tag{$\to\rm{def}$}\\ &=\frac{1}{4}\left(q\left(\psi+\phi\right)-q\left(\psi-\phi\right) +i q\left(\psi+i \phi\right)-i q\left(\psi-i\phi\right)\right)\\ &\,\,=\color{blue}{\tilde{q}(\psi,\phi)} \end{align*} and (4.2) follows. We now obtain \begin{align*} \color{blue}{\tilde{q}\left(\phi+\psi,\xi\right)}&=\tilde{q}\left(\xi,\phi+\psi\right)^{*}\tag{$\to (4.2)$}\\ &=\left(\tilde{q}\left(\xi,\phi\right)+\tilde{q}\left(\xi,\psi\right)\right)^{*}\tag{$\to (3)$}\\ &=\tilde{q}\left(\xi,\phi\right)^{*}+\tilde{q}\left(\xi,\psi\right)^{*}\tag{$\to \rm{def}$}\\ &\,\,\color{blue}{=\tilde{q}\left(\phi,\xi\right)+\tilde{q}\left(\psi,\xi\right)}\tag{$\to (4.2)$}\\ \end{align*} and the claim (4.1), additivity in the second component follows.

conjugate homogeneity second component

We want to show for all $\phi,\psi \in V$ and for all $a\in \mathbb{C}$: \begin{align*} \color{blue}{\tilde{q}\left(\phi,a\psi\right)=a^{*}\tilde{q}\left(\phi,\psi\right)}\tag{5.1} \end{align*} In order to show (5.1) for arbitray $a\in\mathbb{C}$ we proceed in some steps starting with dyadic rationals. We can show \begin{align*} \tilde{q}\left(\phi,\frac{m}{2^n}\psi\right)&=\frac{m}{2^n}\tilde{q}\left(\phi,\psi\right)\qquad\qquad m,n\in\mathbb{N_{0}}\tag{5.2}\\ \tilde{q}\left(\phi,a\psi\right)&=a\tilde{q}\left(\phi,\psi\right)\qquad\qquad\quad\ a\in\mathbb{R_{0}^{+}}\tag{5.3}\\ \tilde{q}\left(\phi,a\psi\right)&=a\tilde{q}\left(\phi,\psi\right)\qquad\qquad\quad\ a\in\mathbb{R}\tag{5.4}\\ \tilde{q}\left(\phi,a\psi\right)&=a^{*}\tilde{q}\left(\phi,\psi\right)\qquad\qquad\quad a\in\mathbb{C}\tag{5.5}\\ \end{align*}

• (5.2): Based upon (2.7) we can show the validity for dyadic rationals by induction.

• (5.3): Since the set of dyadic rational numbers is dense in $\mathbb{R}$, we can use a continuity argument to show the validity of (5.1) for $a\geq 0$. In order to do so we recall that for a norm $p:V\to\mathbb{R}_0^{+}$ we have due to the triangle inequality \begin{align*} |p(\phi)-p(\psi)|\leq p\left(\phi\pm\psi\right) \end{align*} It follows for all $\phi,\psi\in V$ and for all $a, a_k\geq 0$ \begin{align*} |p\left(\phi\pm a_k\psi\right)-p\left(\phi\pm a\psi\right)| &\leq p\left(a_k\psi-a\psi\right)\leq \left|a_k-a\right|p(\psi)\\ |p\left(\phi\pm i a_k\psi\right)-p\left(\phi\pm i a\psi\right)| &\leq p\left(i a_k\psi-i a\psi\right)\leq \left|a_k-a\right|p(\psi)\\ \end{align*} which can be used in conjunction with the polarisation identity (1) to show the claim. We use thereby a sequence $(a_k)_{k\geq 0}$ of non-negative dyadic rationals converging to $a\geq 0$.

• (5.4): Here we can use (1) to show $\tilde{q}\left(\phi,-\psi\right)=-\tilde{q}\left(\phi,\psi\right)$ from which (5.4) follows.

• (5.5): Here we can use (1) to show $\tilde{q}\left(\phi,i\psi\right)=-i\tilde{q}\left(\phi,\psi\right)$ from which (5.5) follows.

and the claim (5.1), conjugate homogeneity in the second component follows.

homogeneity first component

We want to show for all $\phi,\psi \in V$ and for all $a\in \mathbb{C}$: \begin{align*} \color{blue}{\tilde{q}\left(a\phi,\psi\right)=a\tilde{q}\left(\phi,\psi\right)}\tag{6} \end{align*}

We obtain \begin{align*} \color{blue}{\tilde{q}\left(a\phi,\psi\right)}&=\tilde{q}\left(\psi,a\phi\right)^{*}\tag{$\to(4.2)$}\\ &=\left(a^{*}\tilde{q}\left(\psi,\phi\right)\right)^{*}\tag{$\to(5.1)$}\\ &=a\tilde{q}\left(\psi,\phi\right)^{*}\tag{$\to \rm{def}$}\\ &\,\,\color{blue}{=a\tilde{q}\left(\phi,\psi\right)}\tag{$\to (4.2)$}\\ \end{align*} and the claim (6), homogeneity in the first component follows.

Conclusion: We have shown $\tilde{q}:V\times V \to \mathbb{C}$ fulfills additivity in both components, homogeneity in the first component and conjugate homogeneity in the second component, so that (*) and (**) follows. This shows that mappings $\tilde{q}$ for which the polarisation identity is valid are sesquilinear forms.